%I #14 Feb 01 2023 12:28:37
%S 81,101,271,303,361,405,505,509,567,653,707,743,813,839,909,1033,1083,
%T 1187,1355,1447,1515,1527,1539,1753,1805,1897,1919,1959,2025,2121,
%U 2229,2381,2439,2511,2517,2525,2527,2545,2579,2687,2727,2749,2753,2777,2803,2835
%N Moduli n for which A248218(n) = 9.
%C If x is a member of this sequence, and y is a member of this sequence or A248219 or A256343, then LCM(x,y) is a member of this sequence. - _Robert Israel_, Mar 09 2021
%H Robert Israel, <a href="/A256349/b256349.txt">Table of n, a(n) for n = 1..10000</a>
%e In Z/81Z, the iteration of x -> x^2+1 starting at x = 0 yields (0, 1, 2, 5, 26, 29, 32, 53, 56, 59, 80, 2, ...), and m = 81 is the least positive number for which there is such a cycle of length 9, here [2, 5, 26, 29, 32, 53, 56, 59, 80], therefore a(1) = 81.
%p filter:= proc(n) local x, k, R,p;
%p x:= 0; R[0]:= 0;
%p for k from 1 do
%p x:= x^2+1 mod n;
%p if assigned(R[x]) then return evalb(k-R[x] = 9)
%p else R[x]:= k
%p fi
%p od;
%p end proc:
%p select(filter, [$1..10000]); # _Robert Israel_, Mar 09 2021
%t filterQ[n_] := Module[{x, k, R}, x = 0; R[0] = 0; For[k = 1, True, k++, x = Mod[x^2 + 1, n]; If[IntegerQ[R[x]], Return[k - R[x] == 9], R[x] = k]]];
%t Select[Range[10000], filterQ] (* _Jean-François Alcover_, Feb 01 2023, after _Robert Israel_ *)
%o (PARI) for(i=1,3000,A248218(i)==9&&print1(i","))
%Y Cf. A248218, A248219, A256342 - A256348, A003095, A247981.
%K nonn
%O 1,1
%A _M. F. Hasler_, Mar 25 2015